[Edit:  I assumed it was the sum of 5 edges that was equal to 3, but instead each of the 5 edges is of length 3, so this post is not relevant to the problem]

I calculate that a regular tetrahedron inscribed in a sphere of radius 2 to be 8(√6) = about 19.596

Our problem has the sum of 5 edges being 3, so this tetrahedron is much smaller than the regular tetrahedron.

Upper limit

Suppose the first 3 points are very close together and the 4th point is about 1.5 units away from the first 3.  So the sixth side would be 1.5.

If the first 3 form an equilateral triangle of side length ε, the 4th point is (1.5 - 1.5ε) distance from the first 3.

Alternate setup.  Call the first 3 points A,B,C and say they are equidistant from each other with edge length x.  D is between them and equidistant from each of them, which if this were a flat plane would be (√3/3)x. So 3 edges of length x and 3 edges of length (√3/3).  To maximize the 6th edge, say it is one of the longer ones.  So 2x + √3x = 3, x = 3/(2+√3) = 6 - 3√3 = about 0.804

So the upper limit seems to be 1.5

Lower limit

Same alternate setup, but let the 6th edge be one of the smaller ones.

3x + 2√3x/3 = 3,  x = 3/(3 + 2√3/3) = about 0.722; which of course is not accurate due to the curvature of the sphere.

I suspect there are an infinite number of solutions bounded by a lower limit of about 0.7 and an upper limit of 1.5.